Question

calculate the mass of iron in 10kg of iron ore which contains 80%of pure ferric oxide

Answer 1

answer is 5.6 kg coz




in 10kg 80% of fe2o3 is 8 kg ...

so no. of moles of fe2o3 is given weight by molecular weight

so 8000g÷160=50moles




then for one mole of fe2o3 there is 112 g of iron then for 50moles it is

50 ×112=5600g=5.6 kg

Answer 2

Pure Fe2O3 =80/100*10

=8 kg

Molecular mass =56*2+16*3

=160a.m.u

160 a.m.u of Fe2O3 weight 8 kg

So,

Molecular mass of Fe=56*2

=112

112 a.m.u of Fe weights =8/160*112

=5.6 kg

Hope it helps....